Question

integral por substituição trigonométrica exercicios resolvidos

Answer 1

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$\displaystyle\sf\int\dfrac{dx}{x^2~\sqrt{16-x^2}}\\\underline{\rm fac_{\!\!,}a}\\\tt x=4~sen~\theta\implies dx=4~cos~\theta~d\theta\\\tt x^2=16sen^2~\theta\\\tt\sqrt{16-x^2}=\sqrt{16-16sen^2\theta}\\\tt\sqrt{16-x^2}=\sqrt{16(1-sen^2\theta)}=\sqrt{16cos^2~\theta}=4cos~\theta$

$\displaystyle\sf\int\dfrac{dx}{x^2~\sqrt{16-x^2}}=\int\dfrac{\diagup\!\!\!\!4\diagup\!\!\!\!\!cos~\theta~d\theta}{16sen^2~\theta\cdot\diagup\!\!\!\!4\diagup\!\!\!\!cos~\theta}=\dfrac{1}{16}\int cosec^2~\theta~d\theta\\\sf-\dfrac{1}{16}cotg~\theta+k\\\underline{\rm usando~o~tri\hat angulo~auxiliar~temos:}\\\rm cotg~\theta=\dfrac{\sqrt{16-x^2}}{x}\\\boxed{\displaystyle\sf\int\dfrac{dx}{x^2~\sqrt{16-x^2}}=-\dfrac{\sqrt{16-x^2}}{16x}+k}$